Building Mathematics

Building Mathematics
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TABLE OF CONTENTS
Table of Contents.............................................................................................................................1
TASK 1............................................................................................................................................2
Scenario 1....................................................................................................................................2
Scenario 2....................................................................................................................................2
Scenario 3....................................................................................................................................3
TASK 2............................................................................................................................................6
Scenario 1....................................................................................................................................6
Scenario 2..................................................................................................................................10
TASK 3..........................................................................................................................................15
Scenario 1..................................................................................................................................15
Scenario 2..................................................................................................................................18
TASK 4..........................................................................................................................................21
Scenario 1..................................................................................................................................21
Scenario 2..................................................................................................................................23
Scenario 3..................................................................................................................................25
Scenario 4..................................................................................................................................26
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TASK 1
Scenario 1
(a) Length and width of rectangle
Solution
Area = Lb = 26.5 m2
Where L is length and b is width
Given b = L – 3.2
L*(L – 3.2) = 26.5
L2 – 3.2L -265 = 0
L = { - (-3.2) +- √ [(-3.2)2- 4*1* (26.5)] } / 2*1
L = 6.99 and l = -3.79
AS L cannot be negative so neglecting negative value
Length = 6.990 m
b = 6.99-3.2 = 3.790 m
(a) Daily Forfeit
Solution
Let forfeit amount of one day is = £v and original amount of contract = £u
If there is delay of 5 days, then total amount which will be deducted from contract amount is £5v
Amount received by contract if there is delay of 5 days is expressed as:
u – 5v = £4250
Similarly, if there is 12 day delay then forfeit amount for 12 days = £12v
Amount received by contract:
u – 12v = £2120
On solving both equations:
u – 5v = £4250 (i)
u – 12v = £2120 (ii)
On subtracting both the equations we have:
7v = 2130
v = £304.28
On substituting this value in equation 1
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u – 5(304.28) = £4250
u = £5771.4
Contract amount u = £5771.4
Forfeit v = £304.28
Scenario 2
(a) Conversions
Solution
Speed = 65 miles per hour
(i) Speed in m/s :
1 mile = 1609.344 meter
1 hour = 3600 seconds so for converting mile/hour into meter/ second * by (1609.344/3600)
65 * 0.44704 = 29.05 meter per second
(ii) Time to cover distance of 100 miles:
Solution
Time = distance/ speed
= 100/65 = 1 hour 53 minutes
For covering distance of 100 miles nearly 1 hour and 53 minutes will be needed.
(iii) Average fuel consumption = 30 miles per gallon
Solution
30 miles per gallon to litre per kilometre:
1 Mile per gallon = 2.35 litre per km
30 miles/ gallon = 30 * 2.35 = 70.5 litre / km
(iv) Fuel required for the journey in litters
Solution
Fuel required for 1 km = 70.5 litre
Total journey 100 miles = 160 km
So fuel required for journey = 11280 litre
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(a) Unit of lift
Solution
Lift = k *V2 * A * p
A = area = m2 = [M0L2T0]
p = air density = kg / m3 = [M1L-3T0]
k is dimensionless as its constant so = [M0L0T0]
V is speed in m/second = [M0L1T-1]
So V2 = [M0L2T-2]
Lift = [M0L2T0] * [M1L-3T0] * [M0L0T0] * [M0L2T-2]
Lift = [M1L1T-2] = kg meter/ second2
Scenario 3
(1) Arithmetic sequence
Solution
b , 2b/3 , b/3 , 0 .....
Sixth term:
First term a = b
Common difference d = 2b/3 – b = - b/3
Sixth term = a + (n-1) *d
= b + 5(-b/3)
= b – 5b/3
= -2b/3
Sixth term = -2b/3
Kth term : a + (k -1) *d where a is first term and d is common difference
For given AP
Kth term = b + (k -1) *(-b/3)
20th value = 15
20th value can be find as:
b + 19 *(-b/3) = 15
so b = -2.81
Sum of first 20 values:
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Sum = (n/2) * [2a + (n-1)*d]
On substituting values:
Sum = (20/2) * [2*(-2.81) + (20-1)*(-2.8)/3]
= 10 [-5.62 – 19b/3]
= 10/3 [-16.86 + 53.39]
= 10/3 *36.53
= 121.76
Sum of 20 terms = 121.76
(2) Geometric progression
Solution
1, 1/2 , 1/4 ....
a = 1
Formula for nth term = a * r (n-1)
n = 20 and r = 0.5
20th term = 1 * 0.5 (20-1)
20th term = 1/ 524288
(2.) Value of sum
The sum of GP up to infinite number of terms is given by formula: S = a / (1-r)
On substituting values: a = 1 and r = 0.5
Sum = 1 / (1-0.5) = 2
Thus sum is equals to 2
(3) Equations
a.) 2 log 3x + log 18x = 27
Solution
Log 18x = log (3x*6)
2 log 3x + log (3x*6) = 27
2 log 3x + log 3x + log 6 = 27
2 log 3x + log 3x = 26.22
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